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    Hi,

    I understand quadratic equations using factoring when the first factor has a coefficient of 1. e.g.

    x^2 +10x + 25

    I get confused when the coefficient is more than 1? e.g.

    6x^2 - 8x - 8

    or

    2x^4 + 14x^2 + 24

    I think its because both equations have a common factor?
    Also, can someone explain decomposition?

    Thank you
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    (Original post by x-Natalie-x)
    Hi,

    I understand quadratic equations using factoring when the first factor has a coefficient of 1. e.g.

    x^2 +10x + 25

    I get confused when the coefficient is more than 1? e.g.

    6x^2 - 8x - 8

    or

    2x^4 + 14x^2 + 24

    I think its because both equations have a common factor?
    Also, can someone explain decomposition?

    Thank you
    Just take out whatever you need

    Eg in your first example, 6x^2 - 8x - 8 can become 6(x^2 -(8x/6) - (8/6)) and factorise the inside and hen multiply the result by 6 at the end.
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    (Original post by x-Natalie-x)
    Hi,

    I understand quadratic equations using factoring when the first factor has a coefficient of 1. e.g.

    x^2 +10x + 25

    I get confused when the coefficient is more than 1? e.g.

    6x^2 - 8x - 8

    or

    2x^4 + 14x^2 + 24

    I think its because both equations have a common factor?
    Also, can someone explain decomposition?

    Thank you
    The first thing I'd do is factor out the highest common factor amongst all the coefficients. So for 6x^2-8x-8 I'd factor out 2 to get 2(3x^2-4x-4)

    Then consider factorising 3x^2-4x-4 and the way you do it is say that it is equal to 3(x+\frac{\lambda}{3})(x+\frac{ \mu}{3}) where \lambda + \mu = -4 and \lambda\mu = -12. One such case is \lambda = -6 and \mu=2 so just sub those back up in there to get the total factorisation of 2(x-2)(3x+2)


    In general, when you have ax^2+bx+c, where a,b,c have no more common factors, it can be factorised into the form \displaystyle a \left( x+\frac{\lambda}{a} \right) \left( x+\frac{\mu}{a} \right) where \lambda+\mu = b and \lambda\mu = ac
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    (Original post by RDKGames)
    The first thing I'd do is factor out the highest common factor amongst all the coefficients. So for 6x^2-8x-8 I'd factor out 2 to get 2(3x^2-4x-4)

    Then consider factorising 3x^2-4x-4 and the way you do it is say that it is equal to 3(x+\frac{\lambda}{3})(x+\frac{ \mu}{3}) where \lambda + \mu = -4 and \lambda\mu = -12. One such case is \lambda = -6 and \mu=2 so just sub those back up in there to get the total factorisation of 3(x-2)(3x+2)


    In general, when you have ax^2+bx+c, where a,b,c have no more common factors, it can be factorised into the form \displaystyle a \left( x+\frac{\lambda}{a} \right) \left( x+\frac{\mu}{a} \right) where \lambda+\mu = b and \lambda\mu = ac
    Out of interest, were you taught this method originally? It's just as good as other methods but I don't think many students learn it in school for some reason.

    I've seen it thought of as turning the original quadratic into a disguised quadratic by multiplying it by a.

    \displaystyle 3x^2-4x-4 = \frac{1}{3}\left((3x)^2-4\cdot(3x) - 12 \right) = \frac{1}{3}\left(y^2-4y-12\right) = ...

    But this is just another way of thinking about exactly the same thing.
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    (Original post by Notnek)
    Out of interest, were you taught this method originally? It's just as good as other methods but I don't think many students learn it in school for some reason.

    I've seen it thought of as turning the original quadratic into a disguised quadratic by multiplying it by a.

    \displaystyle 3x^2-4x-4 = \frac{1}{3}\left((3x)^2-4\cdot(3x) - 12 \right) = \frac{1}{3}\left(y^2-4y-12\right) = ...

    But this is just another way of thinking about exactly the same thing.
    Don't remember being taught a concrete, simple, and easy to remember method in school to be honest. For this, I just googled some simple method and stuck to it.

    Though I do prefer your method as it's much faster in practice. I don't factorise quadratics as much anymore to have looked for any simpler method aha
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    (Original post by RDKGames)
    Don't remember being taught a concrete, simple, and easy to remember method in school to be honest. For this, I just googled some simple method and stuck to it.

    Though I do prefer your method as it's much faster in practice. I don't factorise quadratics as much anymore to have looked for any simpler method aha
    Haha yes you don't really have to factorise many quadratics in a maths degree

    My method has always been to write out two brackets e.g. (3x + __)(x + __) and work out the missing numbers. But this takes practice so I understand why a lot of students don't like it.
 
 
 
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